# How do you solve for x in: ln(3)=ln(10)-(0.1)(x)?

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Solve just like any equation:

ln(3) = ln(10) - (0.1)(x)

-0.1x = [ln(3) - ln(10)]

x = [ln(3) - ln(10)] / -0.1 = 12.03972804

• stanschim
Lv 7
1 month agoReport

I simply subtracted ln(10) from each side and wrote the equation with the terms reversed from the original problem.  Remember, try to isolate the unknown, x, on the left side of the equation when doing algebra.

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• ln(3) = ln(10) - (0.1)x, ie. ln(10) - ln(3) = ln(10/3) = (0.1)x. Multiply through

bt 10 getting x = 10ln(10/3) = 10(1.203972804) = 12.0397804.

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• ln(3)=ln(10)-(0.1)(x)?

0.1x = ln 10 -ln 3=ln 10/3

10/3= e^(0.1x) =e^(x/10)

x = 10 ln(10/3)=12.04

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• ln(3) =  ln(10) - (0.1)(x) ⇒

0.1x = ln 10 - ln 3

.......ln 10 - ln 3

x = ---------------

............0.1

......2.30259 - 1.09861

x = ------------------------

............0.1

x = 12.0398......................ANS

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